Question

Знайдіть котангенс кута між діагоналлю куба та його площиною однієї з його граней

Answer 1

Пусть сторона куба x см, 
AB=x
BD=x√2
$B_{1}D=x \sqrt{3}$
$tg \ \textless \ B_{1}DB= \frac{B_{1}B}{BD}= \frac{x}{x \sqrt{2} } = \frac{1}{ \sqrt{2} }$
$B_{1}DB=arctg \frac{1}{ \sqrt{2} }$

Answer 2

Finding the Cotangent of the Angle Between the Cube's Diagonal and a Face

To find the cotangent of the angle between the cube's diagonal and one of its faces, we can use the formula:

cot(θ) = 1 / tan(θ)

Where θ is the angle between the cube's diagonal and one of its faces.

Given that the tangent of an angle is equal to the opposite side divided by the adjacent side, we can use the cube's properties to calculate the cotangent.

Let's denote: - d as the length of the cube's diagonal - s as the length of one side of the cube

The cotangent of the angle between the cube's diagonal and one of its faces can be calculated using the formula:

cot(θ) = 1 / (d / s)

Now, let's substitute the values of the cube's diagonal and the length of its side into the formula to find the cotangent of the angle.

Using the provided search result, we have the following values: - Length of the cube's diagonal (d): 4 - Length of one side of the cube (s): 7

Substituting these values into the formula: cot(θ) = 1 / (4 / 7) = 7 / 4

Therefore, the cotangent of the angle between the cube's diagonal and one of its faces is 7/4.